Since $(x+1)^3-x^3=3x^2+3x+1$ and $x^3-(x+1)^3=-3x^2-3x-1$, to say that the difference of two cubes is divisible by 5 is the same as saying that $3x^2+3x+1\equiv 0\mod 5$ or $-3x^2-3x-1\equiv 0\mod 5$. Both of these statements imply that $x(x+1)\equiv 3\mod 5$. Thus I can finish this by showing that there are no such integers which satisfy $x(x+1)\equiv 3\mod 5$.

I want to say that it is sufficient to check by hand for the values 0,1,2,3, and 4 (for which it is not true), but other than following this by a messy induction I was wondering if there is an easier way to show that there are no integers such that $x(x+1)\equiv 3\mod 5$?

Since you are considering things like $(x+1)^3-x^3$, I assume the problem should be to show that the difference of two consecutive cubes is never divisible by $5$.

You could finish the problem in this way:
$$\eqalign{
x(x+1)\equiv3\pmod5\quad
&\Rightarrow\quad 4x(x+1)+1\equiv 3\pmod5\cr
&\Rightarrow\quad (2x+1)^2\equiv 3\pmod5\cr}$$
which implies that $3$ is a square (the technical term is "quadratic residue") modulo $5$. But it isn't, so this is impossible.

How do you know that $3$ is not a square modulo $5$?

Method $1$: calculate $0^2,1^2,\ldots,4^2$ modulo $5$ and observe that you never get $3$. This is no easier than what you suggested!

Method $2$: use the Legendre symbol and quadratic reciprocity. This is probably more work than what you suggested, though if you had a larger modulus instead of $5$ it would probably be better.

If you've already checked 0,1,2,3,4 by hand, you don't have to do any induction, because $x(x+1) \equiv y(y+1)$ for some $y \in [0..4]$. Likewise, couldn't you do exactly the same thing on the original question? Cubes mod 5 are just as easy to compute as squares mod 5.